physics
posted by Becky .
A daring ranch hand sitting on a tree limb wishes to drop vertically onto a horse galloping under the tree. The constant speed of the horse is 15.0 m/s, and the distance from the limb to the level of the saddle is 3.50 m.
(a) What must be the horizontal distance between the saddle and limb when the ranch hand makes his move?
(b) How long is he in the air?
I used the kinematic equations of motion but I can't seem to get an answer that makes sense.

He needs to fall 3.5m
time in air: 3.5=1/2 g t^2
now having the time in air,
distance= 15*t 
To solve for the first part, I got time = 2.89 seconds, but then it shows that I got the wrong answer?

Nah bobpursely is definitely wrong. Anyone know how to solve this?

Yea the solution provided is incorrect.

The correct solution is this:
3.5/4.9 = some answer s^2
then remember
someanswer s^2 = t^2
so solve for t to get .845 seconds for time in air.